Answer :
[tex]\tan ^2x+\sec x=1[/tex]
Given the next trigonometric identity:
[tex]\begin{gathered} \tan ^2x+1=\sec ^2x \\ \text{ Or} \\ \tan ^2x=\sec ^2x-1 \end{gathered}[/tex]Substituting this identity into the equation:
[tex]\sec ^2x-1+\sec x=1[/tex]Subtracting 1 at both sides of the equation:
[tex]\begin{gathered} \sec ^2x-1+\sec x-1=1-1 \\ \sec ^2x+\sec x-2=0 \end{gathered}[/tex]Replacing with:
[tex]\begin{gathered} y=\sec x \\ y^2=\sec ^2x \end{gathered}[/tex]we get:
[tex]y^2+y-2=0[/tex]Applying the quadratic formula with a = 1, b = 1 and c = -2:
[tex]\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}[/tex]Recalling that y = sec(x), then we have two options:
[tex]\begin{gathered} \sec x=1 \\ \text{and} \\ \sec x=-2 \end{gathered}[/tex]By definition:
[tex]\sec x=\frac{1}{\cos x}[/tex]Therefore, the first option is:
[tex]\begin{gathered} \frac{1}{\cos x}=1 \\ (\frac{1}{\cos x})^{-1}=1^{-1} \\ \cos x=1 \end{gathered}[/tex]In the interval of x [0,2π), the solution to this equation is 0.
Now, considering the second option:
[tex]\begin{gathered} \frac{1}{\cos x}=-2 \\ (\frac{1}{\cos x})^{-1}=(-2)^{-1} \\ \cos x=-\frac{1}{2} \end{gathered}[/tex]In the interval of x [0,2π), the solutions to this equation are 2π/3 and 4π/3.
In summary, the solutions to tan^2(x) + sec(x) = 1 are:
[tex]x=0,\frac{2\pi}{3},\frac{4\pi}{3}[/tex]